Question: $\int3x^2(x^3+1)^6dx\,=$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{x^7}{7}+C$ (Choice B) B $\dfrac{x^3(x^3+1)^7}{7}+C$ (Choice C) C $\dfrac{(x^3+1)^7}{7}+C$ (Choice D) D $\dfrac{x^3\left(\dfrac14x^4+x\right)^7}{7}+C$
Explanation: Notice that we can rewrite the integral as $\int(x^3+1)^6 \cdot 3x^2dx\,$ If we let $ {u=x^3+1}$, then $du=3x^2 \, dx}$. Substituting gives us: $\int({x^3+1})^6 \cdot 3x^2dx}\,= \int {u}^6\, du}\,$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= \int u^6\,du~\\\\\\ &=\dfrac{u^7}{7}+C\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=\dfrac{u^7}{7}+C\\\\\\&=\dfrac{(x^3+1)^7}{7}+C\end{aligned}$ The answer: $\int3x^2(x^3+1)^6dx\,=\dfrac{(x^3+1)^7}{7}+C$